Free Sample Math Olympiads M-Level

Using order of operations, start from the left checking for parentheses and then exponents. Since there are none, search for division and multiplication (since they belong to the same order). Follow the left to right rule, so first divide 16 by 8 to get 2 and then multiply that result by 2 to get 4. Now perform the addition and subtract working from left to right: 15 – 4 + 6 = 11 + 6 = 17.

Notice that successive differences increase by 3. Add the proper next successive difference to the previous pentagonal number to obtain the next pentagonal number. The 9th penthagonal number equals 117.

The shortest path from A to B consists of 3 steps: one right, one up, and one backward. Select any one of the 3 ways to start. Then there are only 2 ways left and then 1 way to complete the path. Therefore, the number of different paths will be 3 × 2 × 1 = 6.

The first decade of life is years 0 to 10 years. The second decade is 10 to 20 years. The third decade is 20 to 30 years and so on. That makes the seventh decade of life 60 to 70 years. Since Andrew is in the sixth year of the seventh decade he is 65 years old. Working backwards would make the year he was born 1950.

Case I: The only two digit number is 43 [1 solution].
Case II: Three digit numbers
(a) Numbers of the form X43 where X may be any digit 1 through 9 [9 solutions].
(b) Numbers of the form 43X where X can be any digit 0 through 9 [10 solutions].
Case III: Four digit numbers
(a) Numbers of the form 43XY, X and Y may be any digits 0 through 9 [10 × 10 = 100 solutions].
(b) Numbers of the form X43Y, X cannot be 0 but Y can be 0 [9 × 10 = 90 solutions].
(c) Numbers of the form XY43, X cannot be 0 but Y can be 0 but XY cannot be 43 since that was
already counted in case III (a). [9 × 10 – 1 = 89 solutions].
There are no cases of single digit or five digit numbers.
Altogether, there are 1 + 9 + 10 + 100 + 90 + 89 = 299 solutions.